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\(5x+2x\cdot\left(2^3\cdot5-3^2\cdot4\right)+5^2=4^3\)
\(\Rightarrow5x+2x\cdot\left(8\cdot5-9\cdot4\right)+25=64\)
\(\Rightarrow5x+2x\cdot\left(40-36\right)=64-25\)
\(\Rightarrow5x+2x\cdot4=39\)
\(\Rightarrow5x+8x=39\)
\(\Rightarrow x\cdot\left(5+8\right)=39\)
\(\Rightarrow13x=39\)
\(\Rightarrow x=\dfrac{39}{13}\)
\(\Rightarrow x=3\)
Vậy: ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : 2 x : 2 2 = 2 5 nên x = 7.
b) Ta có: 3 x : 3 2 = 3 5 nên x = 7.
c) Ta có : 4 4 : 4 x = 4 2 nên x = 2.
d) Ta có : 5 x : 5 2 = 5 2 nên x = 4,
e) Ta có: 5 x + 1 : 5 = 5 4 nên x = 4.
f) Ta có : 4 2 x - 1 : 4 = 4 2 nên x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(\left(-86-x\right)-\left(3+2x\right)=-4-15\)
\(\Leftrightarrow-86-x-3-2x+4+15=0\)
\(\Leftrightarrow-3x-70=0\)
\(\Leftrightarrow-3x=70\)
hay \(x=-\dfrac{70}{3}\)
Vậy: \(x=-\dfrac{70}{3}\)
2) Ta có: \(18+\left(-x\right)-\left(40-28\right)=-32-\left(-18\right)\)
\(\Leftrightarrow18-x-40+28+32-18=0\)
\(\Leftrightarrow-x+20=0\)
\(\Leftrightarrow-x=-20\)
hay x=20
Vậy: x=20
3) Ta có: \(-27-\left(-31+x\right)-25=-5-17\)
\(\Leftrightarrow-27+31-x-25+5+17=0\)
\(\Leftrightarrow-x+1=0\)
\(\Leftrightarrow-x=-1\)
hay x=1
Vậy: x=1
4) Ta có: \(-9-14-x+42-38=-5+13\)
\(\Leftrightarrow-x-19=8\)
\(\Leftrightarrow-x=27\)
hay x=-27
Vậy: x=-27
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
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\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5\(x\) + 50 = 3\(x\) + 32
5\(x\) + 1 = 3\(x\) + 9
5\(x\) - 3\(x\) = 9 -1
2\(x\) = 8
\(x\) = 8:2
\(x\) = 4
5x + x = 36
6x = 36
x = 36 : 6
x = 6
Vậy x = 6
2 ( x + 4 ) - 32 = 18
2 ( x + 4 ) = 18 + 32
2 ( x + 4 ) = 50
x + 4 = 50 : 2
x + 4 = 25
x = 25 - 4
x = 21
Vậy x = 21
7x - 2x = 50
5x = 50
x = 50 : 5
x = 10
Vậy x = 10
Chúc hok tốt
=>6x=36
X=36:6
X=6
Vậy.......